The gravitational force of attraction between Earth and Venus, if the distance between them is `2.5xx10^(7)` km, is [mass of Venus = `4.8xx1^(24)` kg, mass of the Earth `=6xx10^(24)` kg

Options

A. `4.1xx10^(18)" N"`
B. `2.1xx10^(18)" N"`
C. `3.1xx10^(18)" N"`
D. `5.1xx10(18)" N"`

Answer

To solve the problem of finding the gravitational force of attraction between Earth and Venus, we will use Newton's law of universal gravitation, which states:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

Where:
- \( F \) is the gravitational force,
- \( G \) is the universal gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \),
- \( m_1 \) is the mass of the first object (Earth),
- \( m_2 \) is the mass of the second object (Venus),
- \( r \) is the distance between the centers of the two objects.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Mass of Earth \( m_1 = 6 \times 10^{24} \, \text{kg} \)
- Mass of Venus \( m_2 = 4.8 \times 10^{24} \, \text{kg} \)
- Distance \( r = 2.5 \times 10^{7} \, \text{km} \)

2. **Convert Distance to Meters:**
- Since \( 1 \, \text{km} = 1000 \, \text{m} \), we convert the distance:
\[ r = 2.5 \times 10^{7} \, \text{km} = 2.5 \times 10^{7} \times 10^{3} \, \text{m} = 2.5 \times 10^{10} \, \text{m} \]

3. **Substitute Values into the Gravitational Force Formula:**
\[ F = \frac{(6.674 \times 10^{-11}) \cdot (6 \times 10^{24}) \cdot (4.8 \times 10^{24})}{(2.5 \times 10^{10})^2} \]

4. **Calculate the Denominator:**
\[ (2.5 \times 10^{10})^2 = 6.25 \times 10^{20} \]

5. **Calculate the Numerator:**
- First, calculate \( 6.674 \times 6 \times 4.8 \):
\[ 6.674 \times 6 = 40.044 \]
\[ 40.044 \times 4.8 \approx 192.2112 \]
- Thus, the numerator becomes:
\[ 192.2112 \times 10^{-11} \times 10^{48} = 192.2112 \times 10^{37} \]

6. **Combine the Numerator and Denominator:**
\[ F = \frac{192.2112 \times 10^{37}}{6.25 \times 10^{20}} = 30.75 \times 10^{17} \]

7. **Final Calculation:**
\[ F = 3.075 \times 10^{18} \, \text{N} \]

### Conclusion:
The gravitational force of attraction between Earth and Venus is approximately:
\[ F \approx 3.1 \times 10^{18} \, \text{N} \]